Question

The decomposition of $C{l}_2{O}_7$ at 400 K in gaseous phase to $C{l}_2$ and $O_2$ is 1st order reaction. After 55 sec at 400 K, the pressure of reaction mixture increases from 0.62 to 1.88 atm. Calculate the rate constant of the reaction. Also, calculate the pressure of reaction mixture after 100 seconds:

• A. $1.58\times {10}^{-2},2.33atm$
• B. $1.58\times {10}^{-3},2.33atm$
• C. $15.8\times {10}^{-2},23.3atm$
• D. None of these

Answer 1

Answer: (A)

$1.58\times {10}^{-2},2.33atm$

$C{l}_2{O}_7\left(g\right)\to C{l}_2\left(g\right)+3.5O_2\left(g\right)$

Initial pressure of $C{l}_2{O}_7=0.62$ atm
Equilibrium pressure of $C{l}_2{O}_7=0.62-x$ atm
Equilibrium pressure of $C{l}_2=x$ atm
Equilibrium pressure of $O_2=3.5x$ atm.

Total equilibrium pressure $=0.62-x+x+3.5x=0.62+3.5x=1.88$ atm
$3.5x=1.26$
$x=0.36$ atm

Hence the equilibrium pressure of $C{l}_2{O}_7=0.62-x=0.62-0.36=0.26$ atm.

The rate constant of the reaction $k=\frac{2.303}{t}log\frac{a}{a-x}$

$k=\frac{2.303}{55sec}log\frac{0.62}{0.26}=1.58\times {10}^{-2}/sec$

After 100 sec,
$1.58\times {10}^{-2}=\frac{2.303}{100}log\frac{0.62}{a-x}$

$0.6861=log\frac{0.62}{a-x}$

$4.854=\frac{0.62}{a-x}$

$a-x=0.1277$

$x=a-0.1277=0.62-0.1277=0.4923$

Total equilibrium pressure $=0.62+3.5x=0.62+3.5\left(0.4923\right)=2.33$ atm